A $5.00 kg$ crate is suspended from the end of a short vertical rope of negligible mass. An upward force $F\left(t\right)$ is applied to the ed of the rope, and the height of the crate above its initial position is given by $y\left(t\right)=\left(2 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)t+\left(0.5 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^3$}\right.\right)t^3$. What is the magnitude of the force $F$ when $t=4.00 s$?

Given data :

$$\begin{gathered} m=5 kg \\ y\left(t\right)=\left(2 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)t+\left(0.5 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^3$}\right.\right)t^3 \\ t=4 s \end{gathered}$$

The acceleration of the crate can be given as :

$$\begin{gathered} a=\frac{d^{2}y}{d{t}^2}=\frac{d{2}^{ }}{d{t}^2}\left(2t+0.5t^3\right) \\ =\frac{d}{dt}\left(2+1.5t^2\right) \\ a =3t \cdots \cdots \cdots \cdots \cdots \cdots \cdots \left(1\right) \end{gathered}$$

Put value of $t=4 s$ in equation $\left(1\right)$

$a=12 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$

The equation of motion can be given as :

$F-mg=ma$

Where $F$ is the applied upward force, $mg$ is the weight of crate and $ma$ is the net force.

$$\begin{gathered} F=m\left(g+a\right) \\ =5\left(10+12\right) \\ =5\times 22 \\ F=110 N \end{gathered}$$

Therefore, net upward force acting on crate is $110N$

Hence correct option is 2